\(\int \frac {1}{2+5 x+3 x^2} \, dx\) [81]

Optimal result

Integrand size = 12, antiderivative size = 13 \[ \int \frac {1}{2+5 x+3 x^2} \, dx=-\log (1+x)+\log (2+3 x) \]

[Out]

-ln(1+x)+ln(2+3*x)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {630, 31} \[ \int \frac {1}{2+5 x+3 x^2} \, dx=\log (3 x+2)-\log (x+1) \]

[In]

Int[(2 + 5*x + 3*x^2)^(-1),x]

[Out]

-Log[1 + x] + Log[2 + 3*x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 630

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[c/q, Int[1/Simp
[b/2 - q/2 + c*x, x], x], x] - Dist[c/q, Int[1/Simp[b/2 + q/2 + c*x, x], x], x]] /; FreeQ[{a, b, c}, x] && NeQ
[b^2 - 4*a*c, 0] && PosQ[b^2 - 4*a*c] && PerfectSquareQ[b^2 - 4*a*c]

Rubi steps \begin{align*} \text {integral}& = 3 \int \frac {1}{2+3 x} \, dx-3 \int \frac {1}{3+3 x} \, dx \\ & = -\log (1+x)+\log (2+3 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \frac {1}{2+5 x+3 x^2} \, dx=-\log (1+x)+\log (2+3 x) \]

[In]

Integrate[(2 + 5*x + 3*x^2)^(-1),x]

[Out]

-Log[1 + x] + Log[2 + 3*x]

Maple [A] (verified)

Time = 2.10 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.92

method	result	size
parallelrisch	\(-\ln \left (1+x \right )+\ln \left (\frac {2}{3}+x \right )\)	\(12\)
default	\(-\ln \left (1+x \right )+\ln \left (2+3 x \right )\)	\(14\)
norman	\(-\ln \left (1+x \right )+\ln \left (2+3 x \right )\)	\(14\)
risch	\(-\ln \left (1+x \right )+\ln \left (2+3 x \right )\)	\(14\)

[In]

int(1/(3*x^2+5*x+2),x,method=_RETURNVERBOSE)

[Out]

-ln(1+x)+ln(2/3+x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \frac {1}{2+5 x+3 x^2} \, dx=\log \left (3 \, x + 2\right ) - \log \left (x + 1\right ) \]

[In]

integrate(1/(3*x^2+5*x+2),x, algorithm="fricas")

[Out]

log(3*x + 2) - log(x + 1)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.77 \[ \int \frac {1}{2+5 x+3 x^2} \, dx=\log {\left (x + \frac {2}{3} \right )} - \log {\left (x + 1 \right )} \]

[In]

integrate(1/(3*x**2+5*x+2),x)

[Out]

log(x + 2/3) - log(x + 1)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \frac {1}{2+5 x+3 x^2} \, dx=\log \left (3 \, x + 2\right ) - \log \left (x + 1\right ) \]

[In]

integrate(1/(3*x^2+5*x+2),x, algorithm="maxima")

[Out]

log(3*x + 2) - log(x + 1)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.15 \[ \int \frac {1}{2+5 x+3 x^2} \, dx=\log \left ({\left | 3 \, x + 2 \right |}\right ) - \log \left ({\left | x + 1 \right |}\right ) \]

[In]

integrate(1/(3*x^2+5*x+2),x, algorithm="giac")

[Out]

log(abs(3*x + 2)) - log(abs(x + 1))

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.62 \[ \int \frac {1}{2+5 x+3 x^2} \, dx=-2\,\mathrm {atanh}\left (6\,x+5\right ) \]

[In]

int(1/(5*x + 3*x^2 + 2),x)

[Out]

-2*atanh(6*x + 5)